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jsf

Posted on 2010-09-21
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Last Modified: 2013-11-24
I am new to jsf.
Creating sample application of spring with jsf.
I have following code in jsp
<%@ taglib uri="http://java.sun.com/jsf/html" prefix="h" %>
<%@ taglib uri="http://java.sun.com/jsf/core" prefix="f" %>

<html>
<head>
<title>enter your name page</title>
</head>
<body>
<f:view>
<h1>
<h:outputText value="JSF 1.2 Tutorial"/>
</h1>
  <h:form id="UserEntryForm">
    <h:outputText value="Enter Your Name:"/>
    <h:inputText value="#{UserBean.userName}" />
    <h:commandButton type="submit" action="welcome" value="OK" />
   </h:form>
</f:view>
</body>
</html>

when I see view source it is showing  following output
<form id="UserEntryForm" name="UserEntryForm" method="post" action="/inputname.jsf" enctype="application/x-www-form-urlencoded">


I am not gettting how action is coming as  inputname.jsf?



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Question by:Manish
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4 Comments
 
LVL 20

Expert Comment

by:a_b
ID: 33724979
action is coming as  inputname.jsf -> This is so that the form postbacks to itself
0
 
LVL 5

Accepted Solution

by:
afibarra earned 75 total points
ID: 33726589
<h:commandButton type="submit" action="welcome" value="OK" />

You are not specifying an action method. So JSF does a form postback.
0
 
LVL 11

Author Comment

by:Manish
ID: 33726976
so what it should be?
0
 
LVL 20

Assisted Solution

by:a_b
a_b earned 50 total points
ID: 33731435
It should be a method in your backing bean.
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